Evaluate this paper on the derivation of the Born rule

[RockyMarciano]

But isn't standard QT(not interpretational) that in the macroscopic limit the expectation value and the Born rule are equivalent, and one can use either one to derive the other, and therefore eithe one can be used as starting postulate?

 

[vanhees71]

Yes, but they talk in the usual probabilistic language, and I've given this explanation for the classicality of many macroscopic observables in the very beginning of this thread. The short paper is a very nice demonstration of how QT is consistent in describing the object and measurement apparatus, which is macroscopic, quantum mechanically. It does not deny or disprove standard quantum theory and/or Born's rule!

 

[vanhees71]

But isn't standard QT(not interpretational) that in the macroscopic limit the expectation value and the Born rule are equivalent, and one can use either one to derive the other, and therefore eithe one can be used as stariting postulate?

Well, it's not so obvious that under any circumstances all macroscopic systems behave classically, and indeed there are examples, where this is not the case, e.g., superfluidity, BECs, superconductivity, etc. are examples where you have specific quantum behavior of macroscopic systems.

 

[RockyMarciano]

Yes, but they talk in the usual probabilistic language, and I've given this explanation for the classicality of many macroscopic observables in the very beginning of this thread. The short paper is a very nice demonstration of how QT is consistent in describing the object and measurement apparatus, which is macroscopic, quantum mechanically. It does not deny or disprove standard quantum theory and/or Born's rule!

This is my point in that post. Regardless of the shortcomings of the Born rule, I don't think Neumaier's foundational attack to Born's rule from within the theory is possible.

 

[A. Neumaier]

But isn't standard QT(not interpretational) that in the macroscopic limit the expectation value and the Born rule are equivalent, and one can use either one to derive the other, and therefore either one can be used as starting postulate?

They are not equivalent.

One can never derive from a rule that only applies to many repeated measurements anything that applies to a single measurement. Thus Born's rule (which is of the first kind) cannot imply the interpretive rule 1 of the authors of the paper under discussion, which is of the second kind. And to go from the second to the first (where it is possible at all) is a highly nontrivial matter since it requires to define the complete measurement process in terms of statistical mechanics. This is the reason why the work of Allahverdyan, Balian, and Nieuwenhuizen is both very relevant and a lot of detailed work.

but they talk in the usual probabilistic language,

Only because it is difficult to avoid using this language. But they don't regard Born's rule as part of the foundation of quantum theory.

Apparently because they were criticized of circularity, they spelled out in the paper mentioned in #85 and #180 in detail which interpretation they want their work to be seen in.

There they deliberately introduced the distinction q-expectation value, q-probability, q-variance, etc. to emphasize that in the statistical mechanics calculus they don't assume the usual probabilities but only the formal quantities with the same names.

 

[vanhees71]

Sure, in "axiomatic" formulations you can rename things as much as you like without changing the mathematical "universe". When it comes to physics, it's of course a different thing. Then you have to relate the theory to what's done in the lab, and I don't see anything in the short paper that contradicts standard QT in this respect, particularly not Born's rule. To the contrary, finally they show that the QT formalism within their simple example is compatible with what's measured, including the QT description of the macroscopic measurement apparatus + heat bath ("environment").

 

[ftr]

Vanhees71, can you comment in the other thread. I know you have said you have unwatched, but I think what is been discussed is very important. Thank you.

 

[RockyMarciano]

They are not equivalent.

One can never derive from a rule that only applies to many repeated measurements anything that applies to a single measurement.

I don't know why you insist on this. It is a basic assumption in QM and statisitical mechanics that macroscopic objects are made of large numbers of microscopic objects and that according to this single measurements done to macroscopic objects are FAPP equivalent to many repeated measurements of those microscopic constituents and in this sense to their expectation value when its uncertainty tends to zero.

 

[vanhees71]

Yes, and if statistical QT weren't very successful in this, physicists would have tried to find a better theory since 1925.

 

[A. Neumaier]

according to this single measurements done to macroscopic objects are FAPP equivalent to many repeated measurements of those microscopic constituents

No. They aren't. A measurement is something that is actually measured and yields a communicatable numerical result.
Thus many repeated experiments yield many recorded results. But a single macroscopic measurement yields only a single recorded result.

 

[vanhees71]

You should visit the physics introductory lab! SCNR.

 

[RockyMarciano]

No. They aren't. A measurement is something that is actually measured and yields a communicatable numerical result.
Thus many repeated experiments yield many recorded results. But a single macroscopic measurement yields only a single recorded result.

This distinction is irrelevant in the context of the QT formalism and statisitical mechanics, which you claim you are not questioning. Every single textbook on QT assumes that lab single measurements results in practice are useful precisely because one can imply that the single macroscopic measurement is a simultaneous measure of very many microscopic constituents in the same state, and that this is also equivalent to measuring repeatedly a single system.
Nothing about communicable or recorded results is in the QT and statisitcal mechanics math formalism.

 

[A. Neumaier]

the single macroscopic measurement is a simultaneous measure of very many microscopic constituents in the same state,

How does the single number measure many things simultaneously?

The International vocabulary of basic and general terms in metrology (VIM) from ISO, the international organization for standardization, may be regarded as the authoritative document defining the concept of a measurement.

2.1
measurement
process of experimentally obtaining information about the magnitude of a quantity
NOTES
1 Measurement implies a measurement procedure, based on a theoretical model.
2 In practice, measurement presupposes a calibrated measuring system, possibly
subsequently verified.

Please tell me the measurement procedure used to measure each of these many microscopic constituents, and the measurement system with which these astronomically many measurements are carried out.

 

[mikeyork]

Although I haven't attempted to follow all the details of the discussion here, I can't help feeling that it boils down to the issue I raised in post #3: that theoretical probability and statistical relative frequency are not necessarily best described as the same thing.

 

[A. Neumaier]

Although I haven't attempted to follow all the details of the discussion here, I can't help feeling that it boils down to the issue I raised in post #3: that theoretical probability and statistical relative frequency are not necessarily the same thing.

It boils down to the point that theoretical expectation values and means over frequently repeated measurements are not the same thing, and that the former makes sense even for cases where no repetitions are made. But this is a hollow phrase only if one doesn't realize the implications of it for the measurement problem.

 

[mikeyork]

It boils down to the point that theoretical expectation values and means over frequently repeated measurements are not the same thing, and that the former makes sense even for cases where no repetitions are made. But this is a hollow phrase only if one doesn't realize the implications of it for the measurement problem.

Well, a theoretical expectation value is a property of a distribution function is it not? So the issue becomes one of interpretation of the distribution.

 

[A. Neumaier]

a theoretical expectation value is a property of a distribution function is it not?

No. Only for randon variables. But quantum operators are not random variables. There is no a priori reason why $Tr\rho A$ should have anything to do with measurement at all. Thus the connection, if any, must be postulated by an interpretation.

 

[mikeyork]

No. Only for randon variables

So why call it an expectation value? AFAIK, the term is used exclusively for the first moment of a distribution function. Even a deterministic variable has a distribution function -- it's a delta-function.

 

[mikeyork]

But quantum operators are not random variables.

They are not any sort of variable. They are operators.

 

[A. Neumaier]

They are not any sort of variable. They are operators.

Yes, and They behave like random variables only under special circumstances, and like deterministic variables even more rarely.

So why call it an expectation value?

By analogy. Things are often called by names borrowed from a special case or an analogy.

Just as all sorts of formal objects are called vectors, not only little arrows. A state vector in quantum mechanics has nothing to do with a vector, except ithat it shares some of its formal properties.

 

[mikeyork]

In still don't get your point. Every measurable variable has a distribution -- even deterministic variables. As I said, a deterministic variable is distributed with a delta-function. But for any actual measurement the theoretical distribution depends on the prior conditions. The prior conditions may give it a pre-determined value or not, as the case may be. And this is not dependent on frequency counting.

It seems to me the issue is simply one of specifying a rule for calculating the prior distribution function and the Born rule serves this purpose very neatly.

 

[A. Neumaier]

Every measurable variable has a distribution -

But an operator is a priori a mathematical object and not a measurable variable and hence has no distribution. Thus its expectation value is a purely theoretical quantity unless you declare by an interpretation rule how it is related to measurement. The interpretation rule used by the authors of the paper under discussion is the one given by the thermal interpretation, which makes no reference to a distribution.

No distributions are needed (or even sensible) to interpret a single measurement. The measurement gives a result, and that's it. Distributions make sense only if you can repeat the measurement arbitrarily often.

 

[mikeyork]

But an operator is a priori a mathematical object and not a measurable variable and hence has no distribution. Thus its expectation value is a purely theoretical quantity unless you declare by an interpretation rule how it is related to measurement.

I never claimed otherwise. I said any variable has a distribution function. That distribution function is what the Born rule provides.

No distributions are needed (or even sensible) to interpret a single measurement. The measurement gives a result, and that's it. Distributions make sense only if you can repeat the measurement arbitrarily often.

Here is where you have not grasped my point. A distribution function is a theoretical quantity derived from a theoretical concept of probability that applies directly to any single measurement regardless of whether you might (or could) count frequencies.

It seems to me you have fallen into the fallacy of assuming a distribution function implies a statistical distribution. But no actual statistical distribution is required. See my post #3 which was all about not confusing probability with statistics.

 

[A. Neumaier]

I said any variable has a distribution function.

But an operator is not a variable.

A non-normal operator such as $A=p+iq$ has not even in principle a distribution function (since measuring $A$ would mean measuring $p$ and $q$), though its expectation $\langle A\rangle:=Tr \rho A$ is well-defined for every state $\rho$!

 

[mikeyork]

But an operator is not a variable.

Does a measurement measure an operator or a variable? Do we measure $<A>$ or $<a>$.

A non-normal operator such as $A=p+iq$ has not even in principle a distribution function (since measuring $A$ would mean measuring $p$ and $q$), though its expectation $\langle A\rangle:=Tr \rho A$ is well-defined for every state $\rho$!

The expectation $\langle a\rangle$ is a variable, not an operator and the variable $a$, with values given by the eigenvalues of $A$, has a distribution function $|<a|\psi>|^2$ for every state $\psi$. That is the entire content of Born rule. It says nothing about operators; only variables. $<a|\psi>$ is what matters not $A$.

 

[Jilang]

Sorry, perhaps I am missing something, but it seems to me that <A> says as much about A as it does about the density function, given they both appear on the RHS.

 

[mikeyork]

Sorry, perhaps I am missing something, but it seems to me that <A> says as much about A as it does about the density function, given they both appear on the RHS.

What are you referring to? We measure the expectation value of the observable variable $a$ (a scalar) not the operator $A$. An operator means nothing until you operate on something.

 

[A. Neumaier]

Does a measurement measure an operator or a variable? Do we measure $<A>$ or $<a>$.

What is measured depends on the kind of measurement.

In a macroscopic measurement we measure $<A>$, in a Born-type measurement we measure an eigenvalue of $A$, in a photodetector we measure the presence of a photon in a beam.

If we measure temperature we measure a parameter occurring in the density operator.

Thus there are lots of possibilities.

 

[A. Neumaier]

The expectation $\langle a\rangle$ is a variable

No, it is a number.

 

[mikeyork]

In a macroscopic measurement we measure $<A>$

What kind of macroscopic measurement do you have in mind? If we measure anything we get a number $a_{macro}$. I have no idea what $<A>$ means except as shorthand for $<\psi|A|\psi>$ which is $<a>$. You wrote $<A> = Tr \rho A$ but $Tr \rho A = <a>$. As far as I can see, $<A>$ (the "expectation value of an operator") may be a useful mathematical idea but is physically meaningless until you operate on something.

in a Born-type measurement we measure an eigenvalue of $A$

Ok, but my point is that we don't measure $<A>$. And whatever variable we do measure, it has a prior (i.e. conditional on state preparation) distribution function $|<a|\psi>|^2$.